\(\int \sec (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [48]

Optimal result

Integrand size = 39, antiderivative size = 151 \[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 (10 A+7 C) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{4/3} \sin (c+d x)}{40 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/3} \sin (c+d x)}{7 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{7/3} \tan (c+d x)}{10 b d} \] Output:

3/40*(10*A+7*C)*hypergeom([-2/3, 1/2],[1/3],cos(d*x+c)^2)*(b*sec(d*x+c))^( 
4/3)*sin(d*x+c)/d/(sin(d*x+c)^2)^(1/2)+3/7*B*hypergeom([-7/6, 1/2],[-1/6], 
cos(d*x+c)^2)*(b*sec(d*x+c))^(7/3)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)+3/1 
0*C*(b*sec(d*x+c))^(7/3)*tan(d*x+c)/b/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.75 \[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \cot (c+d x) (b \sec (c+d x))^{7/3} \sqrt {-\tan ^2(c+d x)} \left ((10 A+7 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\sec ^2(c+d x)\right )+7 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\sec ^2(c+d x)\right ) \sec (c+d x)-7 C \sqrt {-\tan ^2(c+d x)}\right )}{70 b d} \] Input:

Integrate[Sec[c + d*x]*(b*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x] + C*Sec[ 
c + d*x]^2),x]
 

Output:

(3*Cot[c + d*x]*(b*Sec[c + d*x])^(7/3)*Sqrt[-Tan[c + d*x]^2]*((10*A + 7*C) 
*Hypergeometric2F1[1/2, 7/6, 13/6, Sec[c + d*x]^2] + 7*B*Hypergeometric2F1 
[1/2, 5/3, 8/3, Sec[c + d*x]^2]*Sec[c + d*x] - 7*C*Sqrt[-Tan[c + d*x]^2])) 
/(70*b*d)
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2030, 3042, 4535, 3042, 4259, 3042, 3122, 4534, 3042, 4259, 3042, 3122}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sec (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 2030

\(\displaystyle \frac {\int (b \sec (c+d x))^{7/3} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{b}\)

\(\Big \downarrow \) 4535

\(\displaystyle \frac {\int (b \sec (c+d x))^{7/3} \left (C \sec ^2(c+d x)+A\right )dx+\frac {B \int (b \sec (c+d x))^{10/3}dx}{b}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{10/3}dx}{b}}{b}\)

\(\Big \downarrow \) 4259

\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{10/3}}dx}{b}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{10/3}}dx}{b}}{b}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}}{b}\)

\(\Big \downarrow \) 4534

\(\displaystyle \frac {\frac {1}{10} (10 A+7 C) \int (b \sec (c+d x))^{7/3}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 d}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{10} (10 A+7 C) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 d}}{b}\)

\(\Big \downarrow \) 4259

\(\displaystyle \frac {\frac {1}{10} (10 A+7 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{7/3}}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 d}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{10} (10 A+7 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{7/3}}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 d}}{b}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\frac {3 b (10 A+7 C) \sin (c+d x) (b \sec (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{40 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 d}}{b}\)

Input:

Int[Sec[c + d*x]*(b*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x] + C*Sec[c + d* 
x]^2),x]
 

Output:

((3*b*(10*A + 7*C)*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]*(b*Se 
c[c + d*x])^(4/3)*Sin[c + d*x])/(40*d*Sqrt[Sin[c + d*x]^2]) + (3*B*Hyperge 
ometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(7/3)*Sin[c + 
 d*x])/(7*d*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(7/3)*Tan[c + d* 
x])/(10*d))/b
 

Defintions of rubi rules used

rule 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 4259

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^(n - 1)*((Sin[c + d*x]/b)^(n - 1)   Int[1/(Sin[c + d*x]/b)^n, x]), x] /; 
FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]
 

rule 4534

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) 
)), x] + Simp[(C*m + A*(m + 1))/(m + 1)   Int[(b*Csc[e + f*x])^m, x], x] /; 
 FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]
 

rule 4535

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

Maple [F]

\[\int \sec \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )d x\]

Input:

int(sec(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Output:

int(sec(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Fricas [F]

\[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right ) \,d x } \] Input:

integrate(sec(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), 
x, algorithm="fricas")
 

Output:

integral((C*b*sec(d*x + c)^4 + B*b*sec(d*x + c)^3 + A*b*sec(d*x + c)^2)*(b 
*sec(d*x + c))^(1/3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)*(b*sec(d*x+c))**(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2 
),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right ) \,d x } \] Input:

integrate(sec(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), 
x, algorithm="maxima")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^(4/3)*s 
ec(d*x + c), x)
 

Giac [F]

\[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right ) \,d x } \] Input:

integrate(sec(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), 
x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^(4/3)*s 
ec(d*x + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \frac {{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\cos \left (c+d\,x\right )} \,d x \] Input:

int(((b/cos(c + d*x))^(4/3)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c 
 + d*x),x)
 

Output:

int(((b/cos(c + d*x))^(4/3)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c 
 + d*x), x)
 

Reduce [F]

\[ \int \sec (c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=b^{\frac {4}{3}} \left (\left (\int \sec \left (d x +c \right )^{\frac {13}{3}}d x \right ) c +\left (\int \sec \left (d x +c \right )^{\frac {10}{3}}d x \right ) b +\left (\int \sec \left (d x +c \right )^{\frac {7}{3}}d x \right ) a \right ) \] Input:

int(sec(d*x+c)*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
                                                                                    
                                                                                    
 

Output:

b**(1/3)*b*(int(sec(c + d*x)**(1/3)*sec(c + d*x)**4,x)*c + int(sec(c + d*x 
)**(1/3)*sec(c + d*x)**3,x)*b + int(sec(c + d*x)**(1/3)*sec(c + d*x)**2,x) 
*a)